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This is obvious,since f (E) = c ⊂ V (c), for any neighborhood V (c) of c ∈ R.Example 2 The function f (x) = x is continuous on R. Indeed, for any point x0 ∈ Rwe have |f (x) − f (x0 )| = |x − x0 | < ε provided |x − x0 | < δ = ε.Example 3 The function f (x) = sin x is continuous on R.In fact, for any point x0 ∈ R we havex + x0x − x0 ≤sin| sin x − sin x0 | = 2 cos22 x − x0 ≤ 2 x − x0 = |x − x0 | < ε,≤ 2sin22 provided |x − x0 | < δ = ε.Here we have used the inequality | sin x| ≤ |x| proved in Example 9 of Paragraph d) of Sect. 3.2.2.Example 4 The function f (x) = cos x is continuous on R.4.1 Basic Definitions and Examples153Indeed, as in the preceding example, for any point x0 ∈ R we havex − x0 x + x0| cos x − cos x0 | = −2 sinsin≤22 x − x0 ≤ |x − x0 | < ε,≤ 2sin2 provided |x − x0 | < δ = ε.Example 5 The function f (x) = a x is continuous on R.Indeed by property 3) of the exponential function (see Paragraph d in Sect.
3.2.2,Example 10a), at any point x0 ∈ R we havelim a x = a x0 ,x→x0which, as we now know, is equivalent to the continuity of the function a x at thepoint x0 .Example 6 The function f (x) = loga x is continuous at any point x0 in its domainof definition R+ = {x ∈ R | x > 0}.In fact, by property 3) of the logarithm (see Paragraph d in Sect. 3.2.2, Example10b), at each point x0 ∈ R+ we havelimR+ x→x0loga x = loga x0 ,which is equivalent to the continuity of the function loga x at the point x0 .Now, given ε > 0, let us try to find a neighborhood UR+ (x0 ) of the point x0 so asto have| loga x − loga x0 | < εat each point x ∈ UR+ (x0 ).This inequality is equivalent to the relations−ε < logax< ε.x0For definiteness assume a > 1; then these last relations are equivalent tox0 a −ε < x < x0 a ε .The open interval ]x0 a −ε , x0 a ε [ is the neighborhood of the point x0 that weare seeking.
It is useful to note that this neighborhood depends on both ε and thepoint x0 , a phenomenon that did not occur in Examples 1–4.1544Continuous FunctionsExample 7 Any sequence f : N → R is a function that is continuous on the set Nof natural numbers, since each point of N is isolated.4.1.2 Points of DiscontinuityTo improve our mastery of the concept of continuity, we shall explain what happensto a function in a neighborhood of a point where it is not continuous.Definition 4 If the function f : E → R is not continuous at a point of E, this pointis called a point of discontinuity or simply a discontinuity of f .By constructing the negation of the statement “the function f : E → R is continuous at the point a ∈ E”, we obtain the following expression of the definition of thestatement that a is a point of discontinuity of f :(a ∈ E is a point of discontinuity of f ) := = ∃V f (a) ∀UE (a) ∃x ∈ UE (a) f (x) ∈/ V f (a) .In other words, a ∈ E is a point of discontinuity of the function f : E → R ifthere is a neighborhood V (f (a)) of the value f (a) that the function assumes at asuch that in any neighborhood UE (a) of a in E there is a point x whose image isnot in V (f (a)).In ε–δ-form, this definition has the following appearance:∃ε > 0 ∀δ > 0 ∃x ∈ E |x − a| < δ ∧ f (x) − f (a) ≥ ε .Let us consider some examples.Example 8 The function f (x) = sgn x is constant and hence continuous in theneighborhood of any point a ∈ R that is different from 0.
But in any neighborhood of 0 its oscillation equals 2. Hence 0 is a point of discontinuity for sgn x. Weremark that this function has a left-hand limit limx→−0 sgn x = −1 and a right-handlimit limx→+0 sgn x = 1. However, in the first place, these limits are not the same;and in the second place, neither of them is equal to the value of sgn x at the point 0,namely sgn 0 = 0. This is a direct verification that 0 is a point of discontinuity forthis function.Example 9 The function f (x) = | sgn x| has the limit limx→0 | sgn x| = 1 as x → 0,but f (0) = | sgn 0| = 0, so that limx→0 f (x) = f (0), and 0 is therefore a point ofdiscontinuity of the function.We remark, however, that in this case, if we were to change the value of thefunction at the point 0 and set it equal to 1 there, we would obtain a function that iscontinuous at 0, that is, we would remove the discontinuity.4.1 Basic Definitions and Examples155Fig.
4.1Definition 5 If a point of discontinuity a ∈ E of the function f : E → R is suchthat there exists a continuous function f˜ : E → R such that f |E\a = f˜|E\a , then ais called a removable discontinuity of the function f .Thus a removable discontinuity is characterized by the fact that the limitlimEx→a f (x) = A exists, but A = f (a), and it suffices to set!f (x) for x ∈ E, x = a,f˜(x) =Afor x = a,in order to obtain a function f˜ : E → R that is continuous at a.Example 10 The function!f (x) =sin x1 ,for x = 0,0,for x = 0,is discontinuous at 0. Moreover, it does not even have a limit as x → 0, since, aswas shown Example 5 in Sect.
3.2.1, limx→0 sin x1 does not exist. The graph of thefunction sin x1 is shown in Fig. 4.1.Examples 8, 9 and 10 explain the following terminology.Definition 6 The point a ∈ E is called a discontinuity of first kind for the functionf : E → R if the following limits2 exist:limEx→a−0f (x) =: f (a − 0),limEx→a+0f (x) =: f (a + 0),but at least one of them is not equal to the value f (a) that the function assumes at a.2 If a is a discontinuity, then a must be a limit point of the set E. It may happen, however, that allthe points of E in some neighborhood of a lie on one side of a.
In that case, only one of the limitsin this definition is considered.1564Continuous FunctionsDefinition 7 If a ∈ E is a point of discontinuity of the function f : E → R and atleast one of the two limits in Definition 6 does not exist, then a is called a discontinuity of second kind.Thus what is meant is that every point of discontinuity that is not a discontinuityof first kind is automatically a discontinuity of second kind.Let us present two more classical examples.Example 11 The function!D(x) =1, if x ∈ Q,0, if x ∈ R\Q,is called the Dirichlet function.3This function is discontinuous at every point, and obviously all of its discontinuities are of second kind, since in every interval there are both rational and irrationalnumbers.Example 12 Consider the Riemann function4!R(x) =1n,if x =0,if x ∈ R\Q.mn∈ Q, wheremnis in lowest terms, n ∈ N,We remark that for any point a ∈ R, any bounded neighborhood U (a) of it, and anynumber N ∈ N, the neighborhood U (a) contains only a finite number of rationalnumbers mn , m ∈ Z, n ∈ N, with n < N .By shrinking the neighborhood, one can then assume that the denominators ofall rational numbers in the neighborhood (except possibly for the point a itself ifa ∈ Q) are larger than N .
Thus at any point x ∈ Ů (a) we have |R(x)| < 1/N .We have thereby shown thatlim R(x) = 0x→aat any point a ∈ R\Q. Hence the Riemann function is continuous at any irrationalnumber. At the remaining points, that is, at points x ∈ Q, the function is discontinuous, except at the point x = 0, and all of these discontinuities are discontinuities offirst kind.3 P.G.Dirichlet (1805–1859) – great German mathematician, an analyst who occupied the post ofprofessor ordinarius at Göttingen University after the death of Gauss in 1855.4 B.F. Riemann (1826–1866) – outstanding German mathematician whose ground-breaking workslaid the foundations of whole areas of modern geometry and analysis.4.2 Properties of Continuous Functions1574.2 Properties of Continuous Functions4.2.1 Local PropertiesThe local properties of functions are those that are determined by the behavior ofthe function in an arbitrarily small neighborhood of the point in its domain of definition.Thus, the local properties themselves characterize the behavior of a function inany limiting relation when the argument of the function tends to the point in question.
For example, the continuity of a function at a point of its domain of definitionis obviously a local property.We shall now exhibit the main local properties of continuous functions.Theorem 1 Let f : E → R be a function that is continuous at the point a ∈ E. Thenthe following statements hold.10 The function f : E → R is bounded in some neighborhood UE (a) of a.20 If f (a) = 0, then in some neighborhood UE (a) all the values of the functionhave the same sign as f (a).30 If the function g : UE (a) → R is defined in some neighborhood of a and, like f ,is continuous at a, then the following functions are defined in some neighborhood of a and continuous at a:a) (f + g)(x) := f (x) + g(x),b) (f · g)(x) := f (x) · g(x),(x)c) ( fg )(x) := fg(x)(provided g(a) = 0).40 If the function g : Y → R is continuous at a point b ∈ Y and f is such thatf : E → Y, f (a) = b, and f is continuous at a, then the composite function(g ◦ f ) is defined on E and continuous at a.Proof To prove this theorem it suffices to recall (see Sect.
4.1) that the continuityof the function f or g at a point a of its domain of definition is equivalent to thecondition that the limit of this function exists over the base Ba of neighborhoods ofa and is equal to the value of the function at a: limBa f (x) = f (a), limBa g(x) =g(a).Thus assertions 10 , 20 , and 30 of Theorem 1 follow immediately from the definition of continuity of a function at a point and the corresponding properties of thelimit of a function.(x)The only explanation required is to verify that the ratio fg(x)is actually definedin some neighborhood ŨE (a) of a.
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